The pH of 0.1 , ,M solution of the following salts increases in t… - Vidyadip

MCQ

The $pH$ of $0.1\,\,M$ solution of the following salts increases in the order

A
$NaCl < N{H_4}Cl < NaCN < HCl$
B
$HCl < NaCl < NaCN < N{H_4}Cl$
C
$NaCN < N{H_4}Cl < NaCl < HCl$
✓
$HCl < N{H_4}Cl < NaCl < NaCN$

✓

Answer

Correct option: D.

$HCl < N{H_4}Cl < NaCl < NaCN$

(d) $N{H_4}Cl$ undergoes cationic hydrolysis hence $pH$ is $ > 7$ because the solution due to cationic hydrolysis in acids.

$NaCN $ undergoes anionic hydrolysis hence $ pH $ is $>7.$

$HCl$ is strong acid and $NaCl$ is neutral solution.

Hence the $pH$ of given solutions will increases.

$HCl < NaCl < NaCN < N{H_4}Cl$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 6.2. Equilibrium - II (icon Equilibrium)→6.2. Equilibrium - II (icon Equilibrium) — all question types→Chemistry STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

$1.520\, g$ of the hydroxide of a metal on ignition gave $0.995\, gm$ of oxide. The equivalent weight of metal is

Select the correct order

$(a)$ $CO < C{O_2} < CO_3^{2 - }$ $ \Rightarrow $ Bond length

$(b)$ ${O_2} < {O_3} < O_2^{ - 2}$ $ \Rightarrow $ Bond length

$(c)$ ${N_2} < N_2^ + $ $ \Rightarrow $ Bond energy

At a given temperature, the equilibrium constant for reaction $PC{l_5}_{(g)}$ $\rightleftharpoons$ $PC{l_3}_{(g)} + C{l_2}_{(g)}$ is $2.4 \times {10^{ - 3}}$. At the same temperature, the equilibrium constant for reaction $PC{l_3}_{(g)} + C{l_2}_{(g)}$ $\rightleftharpoons$ $PC{l_5}_{(g)}$ is

The option($s$) in which at least three molecules follow Octet Rule is(are)

$(A)$ $CO _2, C _2 H _4, NO$ and $HCl$

$(B)$ $NO _2, O _3, HCl$ and $H _2 SO _4$

$(C)$ $BCl _3, NO , NO _2$ and $H _2 SO _4$

$(D)$ $CO _2, BCl _3, O _3$ and $C _2 H _4$

Consider the following equilibrium $AgCl\downarrow +2N{{H}_{3}}\rightleftharpoons {{\left[ Ag{{\left( N{{H}_{3}} \right)}_{2}} \right]}^{+}}+C{{l}^{-}}$ White precipitate of $AgCl$ appears on adding which of the following

In $\ce{H_2O_2}$, the degree of hydrogen bonding is:

When $Cu$ or $Zn$ added to $CuSO_4$ we get $Cu$ .It is due to

Given below are two statements:

Statement-$I$: Since fluorine is more electronegative than nitrogen, the net dipole moment of $\mathrm{NF}_3$ is greater than $\mathrm{NH}_3$.

Statement-$II$: In $\mathrm{NH}_3$, the orbital dipole due to lone pair and the dipole moment of $\mathrm{NH}$ bonds are in opposite direction, but in $\mathrm{NF}_3$ the orbital dipole due to lone pair and dipole moments of $N-F$ bonds are in same direction. In the light of the above statements. Choose the most appropriate from the options given below.

When $pH$ of a solution decreases, its hydrogen ion concentration

$STATEMENT$-$1$: Boron always forms covalent bond.

because

$STATEMENT$-$2$: The small size of $\mathrm{B}^{3+}$ favours formation of covalent bond.